Commuting <covariant derivatives> and using $U^b\nabla_bU_a=0$ gives the optical evolution equation
$$
U^c\nabla_cB_{ab}=-B_{ac}B^c{}_b-R_{acbd}U^cU^d.
$$
Taking its screen trace yields
$$
\frac{d\theta}{d\lambda}
=-\widehat B_{ab}\widehat B^{ba}-R_{ab}U^aU^b.
$$
The optical decomposition and the symmetry or antisymmetry of its pieces imply
$$
\widehat B_{ab}\widehat B^{ba}
=\frac12\theta^2+widehat\sigma_{ab}\widehat\sigma^{ab}
-\widehat\omega_{ab}\widehat\omega^{ab}.
$$
Thus the four-dimensional <Null Raychaudhuri equation> is
$$
\boxed{\frac{d\theta}{d\lambda}
=-\frac12\theta^2
-\widehat\sigma_{ab}\widehat\sigma^{ab}
+\widehat\omega_{ab}\widehat\omega^{ab}
-R_{ab}U^aU^b}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-311/1/b/solution